Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 10 Find all points of discontinuity of f, where f is defined by π(π₯)={ β(π₯+1, ππ π₯β₯1@&π₯2+1 , ππ π₯<1)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 1 When x < 1 When x > 1 Case 1 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) ((1ββ)^2+1) = (1β0)^2+1 = 1^2+1 = 1 + 1 = 2 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) (1+β)+1 = limβ¬(hβ0) 2+β = 2 + 0 = 2 & π(1) = π₯+1 = 1+1 = 2 Hence, L.H.L = R.H.L = π(1) β΄ f is continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x2 + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 1 & π(1) = π₯+1 = 1+1 = 2 Hence, L.H.L = R.H.L = π(1) β΄ f is continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x2 + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence, there is no point of discontinuity Thus, f is continuous for all πβπ